Monday, December 26, 2011

Linear Algebra - Solutions

We now look at the previous set of linear algebra problems and their solutions:

Problems:

1) Let V = R^2 and let W be the subspace (2,1). Let U be the subspace generated by (0, 1). Show that V is the direct sum of W and U.

Solution:

We have: U = (u1, u2) = (0, 1) and W = (w1, w2) = (2,1)

By Th. (4): If: U + W = V and if U/\W = {0} then V is the direct sum of U and W.

We see: U/\W = (u1 + w1) - (u2 + w2) = 2 - 2 = {0}

Then: V = U + W = (u1, u2) + (w1, w2) = (0, 1) + (2, 1) = (2, 2)

2) Prove theorems (1) - (4)

Theorem (1): Given V is a vector space and one basis has m elements and another basis has n, then m = n.

Proof: Let w1, w2......wn be elements of V, and assume n > m. Then w1, w2......wn are linearly independent. If we let {v1, v2.......vm} be a basis of V over the field F then there exist elements a1, a2.........am C F such that:

w1 = a1v1 + a2 w2 + ...........an wm

By assumption we know w not = 0, hence some ai not = 0. Now, say a1 not = 0 without loss of generality then solve for v1:

a1 v1 = w1 - a2 v2 - ..........am vm

v1 = w1/ a1 - a2 v2/ a1 - ...............-am vm/ a1

Thus the subspace of V generated by w1, v2..........vm contains v1 and thus must span all of V since v1, v2......vm generate V. This implies both n > m and m > n are impossible, hence m must equal n.

Theorem (2): Let V be a vector space and {v1, v2, v3.......vn } be a maximal set of linearly independent elements of V, then {v1, v2, v3.......vn } is a basis of V.

Proof: We need to show every element of V can be expressed as a linear combination of v1, v2....vn. Let v be an element of V. Then the elements w, v1,......vn of V must be linearly dependent by hypothesis, hence there exist numbers xo, x1, x n not all zero such that:

xo w + x1 v1 +.........................xn vn = 0

We can't have xo = 0 because if that were so, we'd obtain linear dependence among v1, v2....vn. Then, solve for w in terms of v1, v2 etc.

w = - (x1/xo) v1 - (x2/xo)v2 -.............- (x n/xo) vn

Proving that w is a linear combination of v1, v2.......vn and hence that {v1, v2........vn} is a basis.

Theorem (3): Let V be a vector space consisting of n elements. Let W be a subspace which does not consist of zero alone. Then W has a basis and the dimension of W is less than or equal to n.

Proof: Let w1 be a non-zero element of W. If {w1} is not a maximal set of linearly independent elements of W, we can find an element w2 of W such that w1, w2 are linearly independent. Proceeding one element at a time there must be an element m < or = n such that we can find linearly independent elements w1, w2........wn such that:

{w1, w2............wn}

is a maximal set of linearly independent elements of W. (See, e.g. the proof for Th. 1 and note the number of such elements is at most n).

Now, use Theorem (2) and we can conclude that {w1, w2......wn} is a basis for W. Again, note the number of such elements is at most n, so the dimension is n.

Theorem (4): Let V be a vector space over the field F and let U,W be subspaces. If: U + W = V and if U/\W = {0} then V is the direct sum of U and W. (Note: /\ denotes intersection)

Proof: Given v is a member of V- by the first assumption- there exist elements u in U and w in W such that v = u + w. Thus, V is the sum of U and W. To prove it is the direct sum, we need to show that these elements u, 2 are uniquely determined. Suppose there exists u' belonging to U and w' belonging to W such that: v = u' + w'. Then:

3) What is the dimension of the space of 2 x 2 matrices? Give a basis for this space.

We posit the space of 2 x 2 matrices belongs to the vector space V. If V is a vector space thenany two bases for V contain the same number of vectors. The number of vectors in a basis for a vector space V is called the dimension of V.

For a member matrix of R^2 we have the most generic bases (forming each column):

(1) (0)(0) (1)

Thus one basis would be:

(1)(0)

and thus dim(R^2) = 2

4) What is the dimension of the space of m x n matrices? Give a basis for this space.

By Theorem (1) both n > m and m > n are impossible, hence m must equal n. Thus, the dimension of the space of m x n matrices is n. A basis for this space is:

(a1n)(a2n)....(amn)

Another would be:

(a11)(a21)...(a m1)

By use of the most generalized matrix written as (a ij) with i = 1,......m, and j = 1.....n. the reader should be abel to verify the preceding.

About Me

Specialized in space physics and solar physics, developed first astronomy curriculum for Caribbean secondary schools, has written thirteen books - the most recent:Fundamentals of Solar Physics. Also: Modern Physics: Notes, Problems and Solutions;:'Beyond Atheism, Beyond God', Astronomy & Astrophysics: Notes, Problems and Solutions', 'Physics Notes for Advanced Level&#39, Mathematical Excursions in Brane Space, Selected Analyses in Solar Flare Plasma Dynamics; and 'A History of Caribbean Secondary School Astronomy'. It details the background to my development and implementation of the first ever astronomy curriculum for secondary schools in the Caribbean.